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il63 [147K]
3 years ago
15

(Atwood’s Machine): Two masses, 9 kg and 12 kg, are attached by a lightweight cord and suspended over a frictionless pulley. Whe

n released, find the acceleration of the system and the tension in the cord.

Physics
2 answers:
Fittoniya [83]3 years ago
5 0

Answer:

Acceleration = 1.428m/s2

Tension = 102.85N

Explanation:

The detailed solution is attached

USPshnik [31]3 years ago
5 0

Answer:

The acceleration of the system is 1.401 m/s² and

The tension in the cord is 100.902 N

Explanation:

Let the 9 kg mass be m

Let the 12 kg mass be M

By Newton's second law of motion we have

For the 9 kg mass, T - mg = ma and for the 12 kg mass we have T - Mg  = -Ma

Here we took the upward acceleration as positive a of the 9 kg mass and the downward acceleration of the 12 kg mass as -a

Solving for T for the 9 kg mass we have

T = mg + ma

Substituting  the value of T in to the 12 kg mass equation, we have

mg + ma - Mg = -Ma or  a = (\frac{M-m}{M+m} )g therefore the acceleration is

1.401 m/s²

and the tension is T = mg + ma = 9×(9.81+1.401) = 100.902 N

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Explanation:

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sertanlavr [38]

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Answer: Option A

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A 12.5843-gram sample of metal bromide, MBr4, was dissolved and, after reaction with silver nitrate, AgNO3, all of the bromide w
barxatty [35]

Answer:

zirconium

Explanation:

Given, Mass of AgBr(s) = 23.0052 g

Molar mass of AgBr(s) = 187.77 g/mol

The formula for the calculation of moles is shown below:

Moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles\ of\ AgBr= \frac{23.0052\ g}{187.77\ g/mol}

Moles\ of\ AgBr= 0.1225\ mol

The reaction taking place is:

MBr_4+4AgNO_3\rightarrow 4AgBr+M(NO_3)__4

From the reaction,

4 moles of AgBr is produced when 1 mole of MBr_4 undergoes reaction

1 mole of AgBr is produced when 1 / 4 mole of MBr_4 undergoes reaction

0.1225 mole of AgBr is produced when \frac {1}{4}\times 0.1225 mole of MBr_4 undergoes reaction

Moles of MBr_4 got reacted = 0.030625 moles

Mass of the sample taken = 12.5843 g

Let the molar mass of the metal = x g/mol

So, Molar mass of MBr_4 = x + 4 × 79.904 g/mol = 319.616 + x g/mol

Thus,

0.030625 = \frac{12.5843}{319.616 + x}

Solve for x,

we get, x = 91.2999 g/mol

<u>The metal shows +4 oxidation state and has mass of 91.2999 g/mol . The metals is zirconium.</u>

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